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Section 2.5
Question 56
Can you do 56 as a guide for me to do the rest on my own?
Question 56
Can you do 56 as a guide for me to do the rest on my own?
56.
f(x)=2x^3+3x^2+18x+27
given zero: 3i -> so we will also have -3i
factors are (x-3i) and (x+3i)
their product is: (x-3i) (x+3i)=x^2-(3i)^2=x^2-9(-1)=x^2+9
use long division
..........(2x+3
x^2+9| 2x^3+3x^2+18x+27
..........2x^3+0*x^2+18x
......................3x^2+27
.........................3x^2+27
...................................0
last factor is 2x+3 and third zero is
2x+3=0
2x=-3
x=-3/2
so, zeros are: 3i, -3i, -3/2
The number that is being divided (in this case, 2x^3+3x^2+18x+27) is called the dividend,
and the number that it is being divided by (in this case, x^2+9) is called the divisor.
I write it similar to this
View attachment 488
I do it this way
.............(
x^2+9 | 2x^3+3x^2+18x+27